Different ways about Passing 2D array of pointer to function C

I met some problems when passing 2d-array to functions today, later I found some solutions like A and B. I knew that 2d-array is stored in a continuous memory as 1d. So in my opinion, there is no 2 dimensional points in fact, compiler split this continuous numbers and made '2 dimensional' concept.

My question is: in C after I explicitly told compiler the parameter is a 'int**', why it did not pass a '2 dimensional' point?

Someone may point out "in C you did not told compiler the group-length", I think it knew this information because if you t开发者_StackOverflowry to reexplain the 12 elements' plain memory into arr[2][6] it will be error like "void printArr(int _arr[][6], int _columns, int _rows)".

A little confused please help...

int arr[][3]={{1,2,3},{5,6,7},{9,10,11},{13,14,15}};

void printArr(int _arr[][3], int _columns, int _rows){ //----->A
void printArr(int (* _arr)[3], int _columns, int _rows){//---->B
void printArr(int** _arr, int _columns, int _rows){//--------->C
    for(int r=0; r< _rows; r++){
        for (int c=0; c<_columns; c++)
            //printf("%4d,", _arr[r][c]);
            printf("%4d,", *(*(_arr+r)+c));
    }   
}
int main(){
    printArr(arr, columns, rows);
}

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Arrays decay into pointers, that means if you have an array of arrays int arr[x][y] it can decay into a pointer to array int (*arr)[y].

int ** arr therefore would be a version of decayed array of pointers int *arr[x], which is an incompatible type with your array.

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int ** is a pointer to a pointer to int, not a pointer to a 2-dim array.

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Style note: don't use leading underscores for your identifiers; such names are reserved for the implementation.

Solution B explicitly captures what is happening; in the call to printArr, the expression arr is implicitly converted from "4-element array of 3-element array of int" to "pointer to 3-element array of int".

However, in the context of a function parameter declaration, the notation T a[] is synonymous with T *a; therefore, solution A is effectively the same as solution B.

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You are right that the compiler reserves a contiguous block of memory so knowing that you can do something like my example. It also works with 1d arrays, ie: printArr((int*)arr, 1, 10);

void printArr(int *_arr, int _columns, int _rows) {
    for(int r = 0; r < _rows; r++) {
        for (int c = 0; c < _columns; c++) {
            printf("%4d,", _arr[r * _columns + c]);
        }
    }   
}

int arr[][3]={{1,2,3},{5,6,7},{9,10,11},{13,14,15}};
printArr((int*)arr, 3, 4);

I think you were very close with C. The two changes are parameterizing with int* rather than int** and your offset computation was incorrect.